If int x^3 e^{5 x} d x = frac{e^{5 x}}{5^4}[f | vedclass

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Question

(A) Using the formula for integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = x^3$ and $v = e^{5x}$. $\int x^3 e^{5x}

Vedclass · Accepted Answer

$125 x^3 - 75 x^2 + 30 x - 6$

Vedclass · Answer

(A) Using the formula for integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = x^3$ and $v = e^{5x}$. $\int x^3 e^{5x} dx = x^3 \frac{e^{5x}}{5} - \int 3x^2 \frac{e^{5x}}{5} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} \int x^2 e^{5x} dx$. Now,$\int x^2 e^{5x} dx = x^2 \frac{e^{5x}}{5} - \int 2x \frac{e^{5x}}{5} dx = \frac{x^2 e^{5x}}{5} - \frac{2}{5} \int x e^{5x} dx$. And $\int x e^{5x} dx = x \frac{e^{5x}}{5} - \int 1 \frac{e^{5x}}{5} dx = \frac{x e^{5x}}{5} - \frac{e^{5x}}{25}$. Substituting these back: $\int x^3 e^{5x} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} [\frac{x^2 e^{5x}}{5} - \frac{2}{5} (\frac{x e^{5x}}{5} - \frac{e^{5x}}{25})] + C$. $= \frac{x^3 e^{5x}}{5} - \frac{3 x^2 e^{5x}}{25} + \frac{6 x e^{5x}}{125} - \frac{6 e^{5x}}{625} + C$. $= \frac{e^{5x}}{625} [125 x^3 - 75 x^2 + 30 x - 6] + C$. Since $5^4 = 625$,we have $f(x) = 125 x^3 - 75 x^2 + 30 x - 6$.

If $\int x^3 e^{5 x} d x = \frac{e^{5 x}}{5^4}[f(x)] + C$,then $f(x)$ is equal to

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